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A local bank that has 15 branches uses a two-digit code to
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27 Jul 2010, 11:25

2

16

00:00

A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

66%(01:06) correct 34%(01:12) wrong based on 584 sessions

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A local bank that has 15 branches uses a two-digit code to represent each of its branches. The same integer can be used for both digits of a code, and a pair of two-digit numbers that are the reverse of each other (such as 17 and 71) are considered as two separate codes. What is the fewest number of different integers required for the 15 codes?

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27 Jul 2010, 11:44

2

2

zest4mba wrote:

A local bank that has 15 branches uses a two-digit code to represent each of its branches. The same integer can be used for both digits of a code, and a pair of two-digit numbers that are the reverse of each other (such as 17 and 71) are considered as two separate codes. What is the fewest number of different integers required for the 15 codes?

ChoicesA 3

B 4

C 5

D 6

E 7

Consider the code XY. If there are \(n\) digits available then X can take \(n\) values and Y can also take \(n\) values, thus from \(n\) digits we can form \(n^2\) different 2-digit codes: this is the same as from 10 digits (0, 1, 2, 3, ..., 9) we can form 10^2=100 different 2-digit numbers (00, 01, 02, ..., 99).

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21 Sep 2010, 15:19

I'm confused how xy would have n*n possibilities...wouldn't it b n*(n-1) possibilities because you would have to have two different digits? For example 17 and 71 are two different codes but 99 and 99 are the same code. Can someone explain?
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21 Sep 2010, 23:00

1

Michmax3 wrote:

I'm confused how xy would have n*n possibilities...wouldn't it b n*(n-1) possibilities because you would have to have two different digits? For example 17 and 71 are two different codes but 99 and 99 are the same code. Can someone explain?

It's always good to test theoretical thoughts on practice:

How many codes can be formed using 2 digits (\(n=2\)), 0 and 1.:

00;01;10;11.

4=2^2.

Or consider the following: how many 2-digit codes can be formed out of 10 digits (0, 1, 2, 3, ..., 9)?00;01;02;...99.

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22 Sep 2010, 08:13

Michmax3 wrote:

I'm confused how xy would have n*n possibilities...wouldn't it b n*(n-1) possibilities because you would have to have two different digits? For example 17 and 71 are two different codes but 99 and 99 are the same code. Can someone explain?

n * (n-1) is to say that 99 will not be chosen. To choose 99 once we are saying n*n should be the combo
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31 Jan 2011, 16:58

Bunuel wrote:

zest4mba wrote:

A local bank that has 15 branches uses a two-digit code to represent each of its branches. The same integer can be used for both digits of a code, and a pair of two-digit numbers that are the reverse of each other (such as 17 and 71) are considered as two separate codes. What is the fewest number of different integers required for the 15 codes?

ChoicesA 3

B 4

C 5

D 6

E 7

Consider the code XY. If there are \(n\) digits available then X can take \(n\) values and Y can also take \(n\) values, thus from \(n\) digits we can form \(n^2\) different 2-digit codes: this is the same as from 10 digits (0, 1, 2, 3, ..., 9) we can form 10^2=100 different 2-digit numbers (00, 01, 02, ..., 99).

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31 Jan 2011, 17:03

mariyea wrote:

Bunuel wrote:

zest4mba wrote:

A local bank that has 15 branches uses a two-digit code to represent each of its branches. The same integer can be used for both digits of a code, and a pair of two-digit numbers that are the reverse of each other (such as 17 and 71) are considered as two separate codes. What is the fewest number of different integers required for the 15 codes?

ChoicesA 3

B 4

C 5

D 6

E 7

Consider the code XY. If there are \(n\) digits available then X can take \(n\) values and Y can also take \(n\) values, thus from \(n\) digits we can form \(n^2\) different 2-digit codes: this is the same as from 10 digits (0, 1, 2, 3, ..., 9) we can form 10^2=100 different 2-digit numbers (00, 01, 02, ..., 99).

Re: A local bank that has 15 branches uses a two-digit code to r
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26 Mar 2013, 09:35

\(N\)= number of integersWe have \(N\)options for the first one, \(N-1\) options for the second. So a total of \(N(N-1)\) combinations, and we want that \(N(N-1)=15\). Now or you plug in the different options, or you solve \(N^2-N-15=0\); the first way seems faster, so lets try with A) 3 : 3*2=6 No B)4*3=12 No again C)5*4=20 YES

C
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Re: A local bank that has 15 branches uses a two-digit code to
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27 Mar 2013, 23:05

Let the required number of digits be n.Considering the given conditions,i)The first digit can be filled up in n ways.ii)The second digit can be filled up in n ways too.So we will get \(n * n = n^2\) numbers.

\(n^2 \geq 15\)

=>\(n \geq 4\) [since n is an integer.]

So,option B will be the correct answer.-------------------------------------------Please press KUDOS if you like my post._________________

Re: A local bank that has 15 branches uses a two-digit code to
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19 Aug 2014, 07:24

1

More easy solution. Think logically.

Pick any two integer.

Integers: 1 & 2

Code: 11, 12, 21, 22 = 4 Codes

Add one more integer: 3

13, 31, 33, 23, 32 = 5 Codes

Add one more integer: 4

44, 14, 41, 24, 42, 34, 43 = 7 Codes

Total = 16 Codes. Enough. Answer: B

2 integers create 4 codes. we need 15 codes.

zest4mba wrote:

A local bank that has 15 branches uses a two-digit code to represent each of its branches. The same integer can be used for both digits of a code, and a pair of two-digit numbers that are the reverse of each other (such as 17 and 71) are considered as two separate codes. What is the fewest number of different integers required for the 15 codes?

Re: A local bank that has 15 branches uses a two-digit code to
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02 Oct 2014, 07:35

A local bank that has 15 branches uses a two-digit code to represent each of its branches. The same integer can be used for both digits of a code, and a pair of two-digit numbers that are the reverse of each other (such as 17 and 71) are considered as two separate codes. What is the fewest number of different integers required for the 15 codes?

A. 3B. 4C. 5D. 6E. 7

We can write thatXC1 + XP2 = 15Lets take X=33+ 6 = 9

Now lets take X=44 + 12 = 16 - This is close to our answer. Hence B is the answer
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